| a + c |
| b + d |
will be greater than c/d, but less than a/b. To shew this, observe that (mx + ny)/(m + n) must lie between x and y, if x and y be unequal: for if x be the less of the two, it is certainly greater than
| mx + nx |
| m + n |
or than x; and if y be the greater of the two, it is certainly less than
| my + ny |
| m + n |
or than y. It therefore lies between x and y. Now let a/b be x, and let c/d be y: then a = bx, c = dy. Now
| bx + dy |
| b + d |
is something between x and y, as was just proved; therefore
| a + c |
| b + d |
is something between a/b and c/d. Again, since a/b and c/d are respectively equal to ap/bp and cq/dq, and since, as has just been proved,